\(\int \frac {e^x x^3+(324+324 x+e^{e^{4 x}} (324+324 x)) \log (1+e^{e^{4 x}})+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} (-1296 e^{4 x} x^2+e^x x^3+(-1296 e^{4 x} x+2 e^x x^2) \log (x)+e^x x \log ^2(x))}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} (x^3+2 x^2 \log (x)+x \log ^2(x))} \, dx\) [277]

Optimal result

Integrand size = 156, antiderivative size = 22 \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=e^x-\frac {324 \log \left (1+e^{e^{4 x}}\right )}{x+\log (x)} \] Output:

exp(x)-324*ln(exp(exp(4*x))+1)/(x+ln(x))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=e^x-\frac {324 \log \left (1+e^{e^{4 x}}\right )}{x+\log (x)} \] Input:

Integrate[(E^x*x^3 + (324 + 324*x + E^E^(4*x)*(324 + 324*x))*Log[1 + E^E^( 
4*x)] + 2*E^x*x^2*Log[x] + E^x*x*Log[x]^2 + E^E^(4*x)*(-1296*E^(4*x)*x^2 + 
 E^x*x^3 + (-1296*E^(4*x)*x + 2*E^x*x^2)*Log[x] + E^x*x*Log[x]^2))/(x^3 + 
2*x^2*Log[x] + x*Log[x]^2 + E^E^(4*x)*(x^3 + 2*x^2*Log[x] + x*Log[x]^2)),x 
]
 

Output:

E^x - (324*Log[1 + E^E^(4*x)])/(x + Log[x])
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^x x^3+2 e^x x^2 \log (x)+e^{e^{4 x}} \left (e^x x^3-1296 e^{4 x} x^2+\left (2 e^x x^2-1296 e^{4 x} x\right ) \log (x)+e^x x \log ^2(x)\right )+e^x x \log ^2(x)+\left (324 x+e^{e^{4 x}} (324 x+324)+324\right ) \log \left (e^{e^{4 x}}+1\right )}{x^3+2 x^2 \log (x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )+x \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {324 (x+1) \log \left (e^{e^{4 x}}+1\right )}{x}-\frac {e^x (x+\log (x)) \left (-e^{e^{4 x}} x-x+1296 e^{3 x+e^{4 x}}-\left (e^{e^{4 x}}+1\right ) \log (x)\right )}{e^{e^{4 x}}+1}}{(x+\log (x))^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (e^x+\frac {324 (x+1) \log \left (e^{e^{4 x}}+1\right )}{x (x+\log (x))^2}-\frac {1296 e^{4 x+e^{4 x}}}{\left (e^{e^{4 x}}+1\right ) (x+\log (x))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 324 \int \frac {\log \left (1+e^{e^{4 x}}\right )}{(x+\log (x))^2}dx+324 \int \frac {\log \left (1+e^{e^{4 x}}\right )}{x (x+\log (x))^2}dx-1296 \int \frac {e^{4 x+e^{4 x}}}{\left (1+e^{e^{4 x}}\right ) (x+\log (x))}dx+e^x\)

Input:

Int[(E^x*x^3 + (324 + 324*x + E^E^(4*x)*(324 + 324*x))*Log[1 + E^E^(4*x)] 
+ 2*E^x*x^2*Log[x] + E^x*x*Log[x]^2 + E^E^(4*x)*(-1296*E^(4*x)*x^2 + E^x*x 
^3 + (-1296*E^(4*x)*x + 2*E^x*x^2)*Log[x] + E^x*x*Log[x]^2))/(x^3 + 2*x^2* 
Log[x] + x*Log[x]^2 + E^E^(4*x)*(x^3 + 2*x^2*Log[x] + x*Log[x]^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 68.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
risch	\({\mathrm e}^{x}-\frac {324 \ln \left ({\mathrm e}^{{\mathrm e}^{4 x}}+1\right )}{x +\ln \left (x \right )}\)	\(20\)
parallelrisch	\(-\frac {-2 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x} \ln \left (x \right )+648 \ln \left ({\mathrm e}^{{\mathrm e}^{4 x}}+1\right )}{2 \left (x +\ln \left (x \right )\right )}\)	\(31\)

Input:

int((((324*x+324)*exp(exp(4*x))+324*x+324)*ln(exp(exp(4*x))+1)+(x*exp(x)*l 
n(x)^2+(-1296*x*exp(4*x)+2*exp(x)*x^2)*ln(x)-1296*x^2*exp(4*x)+exp(x)*x^3) 
*exp(exp(4*x))+x*exp(x)*ln(x)^2+2*x^2*exp(x)*ln(x)+exp(x)*x^3)/((x*ln(x)^2 
+2*x^2*ln(x)+x^3)*exp(exp(4*x))+x*ln(x)^2+2*x^2*ln(x)+x^3),x,method=_RETUR 
NVERBOSE)
 

Output:

exp(x)-324*ln(exp(exp(4*x))+1)/(x+ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=\frac {x e^{x} + e^{x} \log \left (x\right ) - 324 \, \log \left (e^{\left (e^{\left (4 \, x\right )}\right )} + 1\right )}{x + \log \left (x\right )} \] Input:

integrate((((324*x+324)*exp(exp(4*x))+324*x+324)*log(exp(exp(4*x))+1)+(x*e 
xp(x)*log(x)^2+(-1296*x*exp(4*x)+2*exp(x)*x^2)*log(x)-1296*x^2*exp(4*x)+ex 
p(x)*x^3)*exp(exp(4*x))+x*exp(x)*log(x)^2+2*x^2*exp(x)*log(x)+exp(x)*x^3)/ 
((x*log(x)^2+2*x^2*log(x)+x^3)*exp(exp(4*x))+x*log(x)^2+2*x^2*log(x)+x^3), 
x, algorithm="fricas")
 

Output:

(x*e^x + e^x*log(x) - 324*log(e^(e^(4*x)) + 1))/(x + log(x))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((324*x+324)*exp(exp(4*x))+324*x+324)*ln(exp(exp(4*x))+1)+(x*ex 
p(x)*ln(x)**2+(-1296*x*exp(4*x)+2*exp(x)*x**2)*ln(x)-1296*x**2*exp(4*x)+ex 
p(x)*x**3)*exp(exp(4*x))+x*exp(x)*ln(x)**2+2*x**2*exp(x)*ln(x)+exp(x)*x**3 
)/((x*ln(x)**2+2*x**2*ln(x)+x**3)*exp(exp(4*x))+x*ln(x)**2+2*x**2*ln(x)+x* 
*3),x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=\frac {{\left (x + \log \left (x\right )\right )} e^{x} - 324 \, \log \left (e^{\left (e^{\left (4 \, x\right )}\right )} + 1\right )}{x + \log \left (x\right )} \] Input:

integrate((((324*x+324)*exp(exp(4*x))+324*x+324)*log(exp(exp(4*x))+1)+(x*e 
xp(x)*log(x)^2+(-1296*x*exp(4*x)+2*exp(x)*x^2)*log(x)-1296*x^2*exp(4*x)+ex 
p(x)*x^3)*exp(exp(4*x))+x*exp(x)*log(x)^2+2*x^2*exp(x)*log(x)+exp(x)*x^3)/ 
((x*log(x)^2+2*x^2*log(x)+x^3)*exp(exp(4*x))+x*log(x)^2+2*x^2*log(x)+x^3), 
x, algorithm="maxima")
 

Output:

((x + log(x))*e^x - 324*log(e^(e^(4*x)) + 1))/(x + log(x))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=\frac {x e^{x} + e^{x} \log \left (x\right ) - 324 \, \log \left ({\left (e^{\left (x + e^{\left (4 \, x\right )}\right )} + e^{x}\right )} e^{\left (-x\right )}\right )}{x + \log \left (x\right )} \] Input:

integrate((((324*x+324)*exp(exp(4*x))+324*x+324)*log(exp(exp(4*x))+1)+(x*e 
xp(x)*log(x)^2+(-1296*x*exp(4*x)+2*exp(x)*x^2)*log(x)-1296*x^2*exp(4*x)+ex 
p(x)*x^3)*exp(exp(4*x))+x*exp(x)*log(x)^2+2*x^2*exp(x)*log(x)+exp(x)*x^3)/ 
((x*log(x)^2+2*x^2*log(x)+x^3)*exp(exp(4*x))+x*log(x)^2+2*x^2*log(x)+x^3), 
x, algorithm="giac")
 

Output:

(x*e^x + e^x*log(x) - 324*log((e^(x + e^(4*x)) + e^x)*e^(-x)))/(x + log(x) 
)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=\int \frac {x^3\,{\mathrm {e}}^x+\ln \left ({\mathrm {e}}^{{\mathrm {e}}^{4\,x}}+1\right )\,\left (324\,x+{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (324\,x+324\right )+324\right )+{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (x^3\,{\mathrm {e}}^x-1296\,x^2\,{\mathrm {e}}^{4\,x}-\ln \left (x\right )\,\left (1296\,x\,{\mathrm {e}}^{4\,x}-2\,x^2\,{\mathrm {e}}^x\right )+x\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2\right )+x\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2+2\,x^2\,{\mathrm {e}}^x\,\ln \left (x\right )}{x\,{\ln \left (x\right )}^2+2\,x^2\,\ln \left (x\right )+{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (x^3+2\,x^2\,\ln \left (x\right )+x\,{\ln \left (x\right )}^2\right )+x^3} \,d x \] Input:

int((x^3*exp(x) + log(exp(exp(4*x)) + 1)*(324*x + exp(exp(4*x))*(324*x + 3 
24) + 324) + exp(exp(4*x))*(x^3*exp(x) - 1296*x^2*exp(4*x) - log(x)*(1296* 
x*exp(4*x) - 2*x^2*exp(x)) + x*exp(x)*log(x)^2) + x*exp(x)*log(x)^2 + 2*x^ 
2*exp(x)*log(x))/(x*log(x)^2 + 2*x^2*log(x) + exp(exp(4*x))*(x*log(x)^2 + 
2*x^2*log(x) + x^3) + x^3),x)
 

Output:

int((x^3*exp(x) + log(exp(exp(4*x)) + 1)*(324*x + exp(exp(4*x))*(324*x + 3 
24) + 324) + exp(exp(4*x))*(x^3*exp(x) - 1296*x^2*exp(4*x) - log(x)*(1296* 
x*exp(4*x) - 2*x^2*exp(x)) + x*exp(x)*log(x)^2) + x*exp(x)*log(x)^2 + 2*x^ 
2*exp(x)*log(x))/(x*log(x)^2 + 2*x^2*log(x) + exp(exp(4*x))*(x*log(x)^2 + 
2*x^2*log(x) + x^3) + x^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x x^3+\left (324+324 x+e^{e^{4 x}} (324+324 x)\right ) \log \left (1+e^{e^{4 x}}\right )+2 e^x x^2 \log (x)+e^x x \log ^2(x)+e^{e^{4 x}} \left (-1296 e^{4 x} x^2+e^x x^3+\left (-1296 e^{4 x} x+2 e^x x^2\right ) \log (x)+e^x x \log ^2(x)\right )}{x^3+2 x^2 \log (x)+x \log ^2(x)+e^{e^{4 x}} \left (x^3+2 x^2 \log (x)+x \log ^2(x)\right )} \, dx=\frac {e^{x} \mathrm {log}\left (x \right )+e^{x} x -324 \,\mathrm {log}\left (e^{e^{4 x}}+1\right )}{\mathrm {log}\left (x \right )+x} \] Input:

int((((324*x+324)*exp(exp(4*x))+324*x+324)*log(exp(exp(4*x))+1)+(x*exp(x)* 
log(x)^2+(-1296*x*exp(4*x)+2*exp(x)*x^2)*log(x)-1296*x^2*exp(4*x)+exp(x)*x 
^3)*exp(exp(4*x))+x*exp(x)*log(x)^2+2*x^2*exp(x)*log(x)+exp(x)*x^3)/((x*lo 
g(x)^2+2*x^2*log(x)+x^3)*exp(exp(4*x))+x*log(x)^2+2*x^2*log(x)+x^3),x)
 

Output:

(e**x*log(x) + e**x*x - 324*log(e**(e**(4*x)) + 1))/(log(x) + x)